Question

expand the expression to a polynomial in standard form: (2x - 1)^4

Explanation:

Step1: Recall the binomial theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((2x-1)^{4}\), we have \(a = 2x\), \(b=-1\), and \(n = 4\).

Step2: Calculate each term of the expansion

• For \(k = 0\):

\(\binom{4}{0}(2x)^{4}(-1)^{0}=\frac{4!}{0!4!}\times16x^{4}\times1 = 16x^{4}\)

• For \(k = 1\):

\(\binom{4}{1}(2x)^{3}(-1)^{1}=\frac{4!}{1!3!}\times8x^{3}\times(-1)=4\times8x^{3}\times(-1)=-32x^{3}\)

• For \(k = 2\):

\(\binom{4}{2}(2x)^{2}(-1)^{2}=\frac{4!}{2!2!}\times4x^{2}\times1 = 6\times4x^{2}=24x^{2}\)

• For \(k = 3\):

\(\binom{4}{3}(2x)^{1}(-1)^{3}=\frac{4!}{3!1!}\times2x\times(-1)=4\times2x\times(-1)=-8x\)

• For \(k = 4\):

\(\binom{4}{4}(2x)^{0}(-1)^{4}=\frac{4!}{4!0!}\times1\times1 = 1\)

Step3: Sum up all the terms

\((2x - 1)^{4}=16x^{4}-32x^{3}+24x^{2}-8x + 1\)

Answer:

\(16x^{4}-32x^{3}+24x^{2}-8x + 1\)